Question

If a photon of intensity $I$ falls on a surface at an angle ${60}^{\circ }$ making with it having absorption coefficient $0.4$, then radiation pressure exerted on the surface is-

• A. $\frac{1.2I}{c}$
• B. $\frac{7I}{20c}$
• C. $\frac{1.05I}{c}$
• D. $\frac{0.4I}{c}$

Answer 1

The correct option is A $\frac{1.2I}{c}$


When a parallel beam of radiation falls on a plane surface at an angle $\theta$ with normal to the surface, if the surface partially reflects (at same angle) and absorbs, then force exerted on surface is,

$F=\frac{IA}{c}(1+r)\cos \theta$

$\text{Component of force normal to surface is,}$

$F_{\perp }=F\cos \theta =\frac{IA}{c}(1+r){\cos }^2\theta$

$\therefore \text{Radiation pressure =}\frac{F\cos \theta }{A}=\frac{I(1+r){\cos }^2\theta }{c}$

Given:
$\alpha =0.4\Rightarrow r=1-\alpha =0.6$
$\varphi ={60}^{\circ }\Rightarrow \theta ={90}^{\circ }-\varphi ={30}^{\circ }$

$\therefore P=\frac{I}{c}(1+0.6)\times (\frac{\sqrt{3}}{2}{)}^2=\frac{1.2I}{c}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{D}\right)$ is the correct answer.

Why this question? Caution: Most of the time, student repeat the mistake while putting the value of angle $\theta$. It is the angle between the incident or reflected ray and normal to the surface.